Advertisements
Advertisements
प्रश्न
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`(sin theta-2sin^3theta)/(2cos^3theta -costheta) = tan theta`
Prove that `(sin theta-2sin^3theta)/(2cos^3theta -costheta) = tan theta`
Advertisements
उत्तर
L.H.S = `(sin theta-2sin^3theta)/(2cos^3theta -costheta)`
= `(sintheta(1-2sin^2theta))/(costheta(2cos^2theta-1))`
= `(sinthetaxx(1-2sin^2theta))/(costhetaxx{2(1-sin^2theta)-1})`
= `(sin thetaxx(1-2sin^2theta))/(costhetaxx(1-2sin^2theta))`
= `tantheta`
= R.H.S
APPEARS IN
संबंधित प्रश्न
Evaluate without using trigonometric tables:
`cos^2 26^@ + cos 64^@ sin 26^@ + (tan 36^@)/(cot 54^@)`
Prove the following trigonometric identities.
`cot^2 A cosec^2B - cot^2 B cosec^2 A = cot^2 A - cot^2 B`
Prove the following identities:
`(1 + sin A)/(1 - sin A) = (cosec A + 1)/(cosec A - 1)`
Prove the following identities:
`(sintheta - 2sin^3theta)/(2cos^3theta - costheta) = tantheta`
`1/((1+tan^2 theta)) + 1/((1+ tan^2 theta))`
Prove the following identity :
( 1 + cotθ - cosecθ) ( 1 + tanθ + secθ)
Prove the following identity :
`sqrt((secq - 1)/(secq + 1)) + sqrt((secq + 1)/(secq - 1))` = 2 cosesq
Prove that: `(sin A + cos A)/(sin A - cos A) + (sin A - cos A)/(sin A + cos A) = 2/(sin^2 A - cos^2 A)`.
1 + cot2θ = ?
If `sec θ + tan θ = sqrt(3)`, complete the activity to find the value of sec θ – tan θ.
Activity:
`square = 1 + tan^2θ` ...[Fundamental trigonometric identity]
`square - tan^2θ = 1`
`(sec θ + tan θ) . (sec θ - tan θ) = square`
`sqrt(3) . (sec θ - tan θ) = 1`
`(sec θ - tan θ) = square`
